A non-zero vectors ${{\bf{\vec a}}}$ is such that its projection along the vectors $\frac{{{\bf{\hat i}} + {\bf{\hat j}}}}{{\sqrt 2 }}$ and $\frac{{ - {\bf{\hat i}} + {\bf{\hat j}}}}{{\sqrt 2 }}$ and ${{\bf{\vec k}}}$ are equal, then unit vector along ${{\bf{\vec a}}}$ is

$\frac{{\sqrt 2 }}{{\sqrt 3 }}{\rm{\;}}{\bf{\hat j}} + \frac{{{\bf{\hat k}}}}{{\sqrt 3 }}$

$\frac{{\sqrt 2 {\bf{\hat j}} - {\bf{\hat k}}}}{{\sqrt 3 }}$

$\frac{{{\bf{\hat j}} - \sqrt 2 {\bf{\hat k}}}}{{\sqrt 3 }}$

$\frac{{{\bf{\hat j}} - {\bf{\hat k}}}}{{\sqrt 2 }}$

JEE MATHEMATICS MCQ: DOT PRODUCT OF VECTORS

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A non-zero vectors

{\mathbf{\vec a}}

is such that its projection along the vectors

\frac{{\mathbf{\hat i} + \mathbf{\hat j}}}{{\sqrt 2 }}

and

\frac{{ - \mathbf{\hat i} + \mathbf{\hat j}}}{{\sqrt 2 }}

and

{\mathbf{\vec k}}

are equal, then unit vector along

{\mathbf{\vec a}}

View answer and explanation

Correct option: C

\frac{{\sqrt 2 }}{{\sqrt 3 }}\mathrm{\;}\mathbf{\hat j} + \frac{{\mathbf{\hat k}}}{{\sqrt 3 }}

Option C (

\frac{{\sqrt 2 }}{{\sqrt 3 }}\mathrm{\;}\mathbf{\hat j} + \frac{{\mathbf{\hat k}}}{{\sqrt 3 }}

) is correct because the projections being equal allows us to determine the components of the unit vector along

\mathbf{\vec a}